Limit Continuity And Differentiability Ques 32
- The function $f(x)=\left\{\begin{array}{ll}|x-3|, & x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}, & x<1\end{array}\right.$ is
$(1988,2 M)$
(a) continuous at $x=1$
(b) differentiable at $x=1$
(c) discontinuous at $x=1$
(d) differentiable at $x=3$
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Answer:
Correct Answer: 32.(a,b)
Solution: (a,b) Here, $f(x)=\left\{\begin{array}{cc}|x-3| & x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}, & x<1\end{array}\right.$
$\therefore \quad$ RHL at $x=1, \lim _{h \rightarrow 0}|1+h-3|=2$
LHL at $x=1$,
$ \lim _{h \rightarrow 0} \frac{(1-h)^2}{4}-\frac{3(1-h)}{2}+\frac{13}{4}=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=\frac{14}{4}-\frac{3}{2}=2 $
$\therefore \quad f(x)$ is continuous at $x=1$
Again, $f(x)=\left\{\begin{array}{cc}-(x-3), & 1 \leq x<3 \\ (x-3), & x \geq 3 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}, & x<1\end{array}\right.$
$\therefore \quad f^{\prime}(x)=\left\{\begin{array}{cc}-1, & 1 \leq x<3 \\ 1, & x \geq 3 \\ \frac{x}{2}-\frac{3}{2}, & x<1\end{array}\right.$
$\left.\begin{array}{ccc}\text { RHD at } x=-1 \Rightarrow 1 \\ \text { LHD at } x= \frac{1}{2}-\frac{3}{2}=-1 \Rightarrow -1\end{array}\right]$ differentiable at $x=1$.
Again, $\left.\begin{array}{ccc}\text { RHD at } x=3 \Rightarrow 1 \\ \text { LHD at } x=3 \Rightarrow -1\end{array}\right]$ not differentiable at $x=3$.
BETA
